help
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As per table, we have
Laplace transform of sin(12t) = 12 / (s^2 + 12^2)
invoke the Shift theorem of laplace transforms
ℒ{e^{at}f(t)} = ℒ{f(t - a)} = ℒ{f(t + 0.4)}
so our transform is
ℒ{e^{-0.4 t} sin(12t)} = ℒ{ sin((12 + 0.4)t} = 12 / ( (s + 0.4)^2 + 12^2)........[Ans.]
Or, just do the raw integral. It is especially easy if you write sin(12t) in terms of the imaginary part of e^{i 12t}, then the integration is trivial.
Laplace transform of sin(12t) = 12 / (s^2 + 12^2)
invoke the Shift theorem of laplace transforms
ℒ{e^{at}f(t)} = ℒ{f(t - a)} = ℒ{f(t + 0.4)}
so our transform is
ℒ{e^{-0.4 t} sin(12t)} = ℒ{ sin((12 + 0.4)t} = 12 / ( (s + 0.4)^2 + 12^2)........[Ans.]
Or, just do the raw integral. It is especially easy if you write sin(12t) in terms of the imaginary part of e^{i 12t}, then the integration is trivial.
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The Laplace transform of f(t) = F(s) = Integral from 0 to inf of {f(t) e^(-st)} dt = Integral from 0 to inf of {e^(-st - 0.4t) sin(12t)} dt = e^(-st - 0.4t)/((s + 0.4)^2 + (12)^2) ((-s - 0.4) sin12t - 12 cos12t) = 0 + 12/((s + 0.4)^2 + 144) = 12/((s + 0.4)^2 + 144)). [I have used the fact that e^(-at) with a > 0 dies off to zero as t goes to +infinity and e^(0) = 1. I also used a table of integrals in the prior step to that....
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sin4"(kuadrat)+costa-4